Question: Assume that $C$ is a positively oriented, piecewise smooth, simple, closed curve. Let $R$ be the region enclosed by $C$. Use the circulation form of Green's theorem to rewrite $ \iint_R 4x^2y^3 - 3y^2x^3 \, dA$ as a line integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \oint_C x^2y^4 \, dx + x^3y^3 \, dy$ (Choice B) B $ \oint_C \dfrac{4x^3y^3}{3} \, dx - x^3y^3 \, dy$ (Choice C) C $ \oint_C -x^2y^4 \, dx + x^3y^3 \, dy$ (Choice D) D $ \oint_C x^3y^3 \, dx + \dfrac{4x^3y^3}{3} \, dy$ (Choice E) E Green's theorem is not necessarily applicable.
Solution: Assume we have a two-dimensional vector field $F(x, y) = P(x, y) \hat{\imath} + Q(x, y) \hat{\jmath}$ and a piecewise smooth, simple, closed curve $C$. Let $R$ be the region enclosed by $C$. Then the circulation form of Green's theorem states that we have the equality below: $ \oint_C P \, dx + Q \, dy = \iint_R \left( \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) dA$ Our first step should be to confirm that the given curve is compatible with using Green's theorem. The curve $C$ does satisfy all the conditions of Green's theorem. It is also positively oriented, which means we don't need to take the negative of our result at the end. The next step is to identify the components $P$ and $Q$ of the vector field $F$ over which we're integrating. We're simplifying from a double integral to a line integral, so we need to find $P$ and $Q$ given that we know: $\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = 4x^2y^3 - 3y^2x^3$ There are many pairs of function $P$ and $Q$ that satisfy the condition. Here's one solution: $\begin{aligned} P(x, y) &= x^3y^3 \\ \\ Q(x, y) &= \dfrac{4x^3y^3}{3} \end{aligned}$ [Is there a strategy to finding good pairs of P and Q?] Now that we know $P$ and $Q$, all that's left is to put everything together. Therefore, one simplification of the double integral is: $ \oint_C x^3y^3 \, dx + \dfrac{4x^3y^3}{3} \, dy$